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STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
MCQ
$\mathop {Lim}\limits_{\lambda  \to 0} \,\,{\left( {\int\limits_0^1 {{{(1 + x)}^\lambda }dx} } \right)^{\frac{1}{\lambda }}}$ is equal to
  • A
    $2\, ln\, 2$
  • ✓
    $\frac{4}{e}$
  • C
    $ln\, \frac{4}{e}$
  • D
    $4$
✓

Answer

Correct option: B.
$\frac{4}{e}$
b
$\mathop {Lim}\limits_{\lambda  \to 0} \,\,{\left( {\int\limits_0^1 {{{(1 + x)}^\lambda }dx} } \right)^{\frac{1}{\lambda }}}$ $= \mathop {Lim}\limits_{\lambda  \to 0} \,\,{\left( {\left. {\frac{{{{(1 + x)}^{\lambda  + 1}}}}{{\lambda  + 1}}} \right|_{\,0}^{\,1}} \right)^{\frac{1}{\lambda }}}$ $= \mathop {Lim}\limits_{\lambda  \to 0} \,\,{\left( {\frac{{{2^{\lambda  + 1}} - 1}}{{\lambda  + 1}}} \right)^{\frac{1}{\lambda }}}$   $(1 ^{\infty}\, form)$
$={e^{\mathop {Lim}\limits_{\lambda  \to 0} \;\frac{1}{\lambda }\;\left( {\frac{{{2^{\lambda  + 1}} - 1 - \lambda  - 1}}{{\lambda  + 1}}} \right)}}$ $={e^{\mathop {Lim}\limits_{\lambda  \to 0} \;\left( {\frac{{{2^{\lambda  + 1}} - 2 - \lambda }}{{\lambda (\lambda  + 1)}}} \right)}}$ $= {e^{\mathop {Lim}\limits_{\lambda  \to 0} \;\left( {\frac{{2({2^\lambda } - 1)}}{\lambda }\; - \;1} \right)}}$ $= e^{2 ln 2 - 1}$ $={e^{\ln \,\left( {\frac{4}{e}} \right)}}$ $=\frac{4}{e}$

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